Find the equation of the plane which contains the line of intersection of the planes `x+2y+3z-4=0a n d2x+y-z+5=0` and which is perpendicular to the plane `5x+3y-6z+8=0.`

Explanation

To find the equation of the plane that contains the line of intersection of the given planes and is perpendicular to another plane, we can follow these steps:

### Step 1: Identify the normal vectors of the given planes
The equations of the planes are:
1. Plane 1: \( x + 2y + 3z - 4 = 0 \) → Normal vector \( \mathbf{n_1} = (1, 2, 3) \)
2. Plane 2: \( 2x + y - z + 5 = 0 \) → Normal vector \( \mathbf{n_2} = (2, 1, -1) \)
3. Plane 3: \( 5x + 3y - 6z + 8 = 0 \) → Normal vector \( \mathbf{n_3} = (5, 3, -6) \)

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